ACI Section 11.5.4.3

[slickdeals]

Greetings,
I have a question pertaining to Section 11.5.4.3 of ACI 318-02. It states that when Vs exceeds 2 Vc, then maximum spacing limit shall be d/4. In case an engineer decides to put more shear steel than necessary, does ACI force section 11.5.4.3 into effect? Because the way Vs is defined, it is not the required shear steel but provided shear steel. It almost seems that the code is penalizing you for being conservative, or am I missing something?

 

[JAE]

If Vs is that high in the first place, and you add MORE stirrups, do you really ever have a condition where d/4 would even control? Might be a rare circumstance.

 

[slickdeals]

I actually had a situation close to that. I had to redesign a plaza slab for a contractor who wanted to drive his concrete trucks over (300 psf LL). To add to my woes, I had headroom issues, making my beam only 16" deep. I had a 48" to 54" wide beam x 16" deep. That's where my problem came from. It is rare, but happened to me now.

 

[slickdeals]

Vu = 250 kips
I have a 48" wide x 16" deep beam
Phi Vc = 74 kips (d=14.5")
Av using a spacing of 4" o.c. = 0.809 in2
If I use 4 #4 ties, Av = 1.2 in2. That gives me Vs = 261 kips > 2Vc, and hence I am forced to use a spacing of 3" o.c.

 

[slickdeals]

Correction Using 3 # 4 ties, Av = 1.2 in2.

 

[JAE]

Actually, referring to your original post - doesn't 11.5.4.3 say 4 x sqrt(f'c)?

 

[slickdeals]

Right, 4*sqrt(fc')bd = 2(2*sqrt(fc')bd) = 2Vc

 

[UcfSE]

I would look for some information about shear reinforcing for wide shallow beams. You may have some modifications you should consider. I seem to remember some research suggesting the ACI typical method is unconservative for this case. It's late and I'm at home so I can't say for sure.

 

[canstruct]

You are misinterpretting the ACI Clause, by calculation you required a spacing of 5.18" and max spacing by ACI restrict in these situations where you have excessive shear stresses vs, shear strenghts. As Vs which is (Vu- phi Vc) /phi exceed 4 sq. rt fc'b.d, then spacing as per ACI should be half not that which you calculated, but the maximum spacing which is d/2 or Av. fy/(50 bw)which is 3.4". And i think you have to check for deflection also which might increase your longitudinal steel also to control your creep deflection.

 

[Lion06]

creep deflection is only helped by COMPRESSION steel, not additional tension steel.
Also, you are not required to check deflections unless the depth of your section is less than that given in table 9.5(a) of ACI 318-05.

 

[slickdeals]

Again, what i do not understand is that if you decide to provide more shear steel than necessary, why would you further decrease the spacing. What is the research that this is based on. And yes, I have checked my deflections and have enough top steel for long term deflections. I will try to do some more research for wide shallow beams. Thanks UcFSE.

 

[JAE]

I guess my view is that the Vs described in this section was simply anticipating that it is the required Vs based on Vs = (Vu - [φ]Vc) / [φ] and not the Vs actually provided.

 

[slickdeals]

I am interpreting it the same way as well.....thanks

 

[MarcbSE]

I have always interpreted the Vs in this case to be the "Required" Vs, not what you provide. If the required shear strength due to stirrups exceeds twice the concrete shear capacity alone, then the maximum stirrup spacing is cut in half.

 

[miecz]

Article 11.5.4 of PCA Notes on 318-02 interprets Vs to be the required steel, not the provided steel.

 

[slickdeals]

I have also sent an email to ACI to see if they will fix the definition of Vs, as it does not make sense intuitively. Vs is something that you choose to provide, and not something you get inherently.

 

[cooperDBM]

When discussing this with ACI you should note that this clause has been renumber 11.5.5.3 in ACI 318-05.

I suppose a more accurate expression for this limit would be,

"Where (Vu/phi - Vc) exceeds 4 sqrt(f'c) bwd ..."

Also, Vc isn't necessarily 2 sqrt(f'c) bwd so the limit doesn't always boil down to 2Vc. You're right though that the general idea is that if the required strength is more than twice the concrete component (originally assuming 2 sqrt(f'c) bwd) then the spacing should be halved since the stirrup strength dominates.

If they did want to assume that Vc = 2 sqrt(f'c) bwd for the sake of this limit then it could be simplified to,

"Where Vu exceeds 6 phi sqrt(f'c) bwd ..." (Vu > 3 phi Vc)

 

[Lion06]

or maybe Vs,req'd